Miscellaneous Exercise

Let $x, y \in \mathbb{R}$ such that $f(x) = f(y)$.

• Case 1: $x, y \ge 0$. Then $|x|=x, |y|=y$.
  $\frac{x}{1+x} = \frac{y}{1+y} \implies x(1+y) = y(1+x) \implies x + xy = y + xy \implies x = y$.
• Case 2: $x, y < 0$. Then $|x|=-x, |y|=-y$.
  $\frac{x}{1-x} = \frac{y}{1-y} \implies x(1-y) = y(1-x) \implies x – xy = y – xy \implies x = y$.
• Case 3: $x > 0, y < 0$. Since $f(x) > 0$ and $f(y) < 0$, $f(x) \neq f(y)$. No conflict.

Thus, $f$ is one-one.

Let $y \in (-1, 1)$. We need to find $x$ such that $f(x) = y$.

• If $y \ge 0$: Solve $y = \frac{x}{1+x} \implies y + yx = x \implies x(1-y) = y \implies x = \frac{y}{1-y}$.
• If $y < 0$: Solve $y = \frac{x}{1-x} \implies y - yx = x \implies x(1+y) = y \implies x = \frac{y}{1+y}$.

Since $y \in (-1, 1)$, the denominator is never zero, so $x$ is always a valid real number. Thus, $f$ is onto.

Let $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$.

Taking the cube root on both sides (which is a unique operation for real numbers):

Therefore, $f$ is injective.

• Reflexive: Is $A \subset A$? Yes, every set is a subset of itself. (Reflexive)
• Symmetric: If $A \subset B$, does it imply $B \subset A$? No. Example: $A=\{1\}, B=\{1,2\}$. $A \subset B$ is true, but $B \not\subset A$. (Not Symmetric)
• Transitive: If $A \subset B$ and $B \subset C$, then $A \subset C$. Yes. (Transitive)

For a function from a finite set to itself ($A \to A$), the properties of Injectivity (One-One) and Surjectivity (Onto) are linked.

If such a function is Onto, it implies that every element in the codomain is mapped to. Since the domain and codomain have the same size ($n$), each element in the domain must map to a unique element in the codomain.

Therefore, the function must be One-One and Onto (Bijective).

The number of bijections on a set of $n$ elements is the number of permutations of those elements.

Two functions are equal if $f(a) = g(a)$ for all $a \in A$. Let’s check each element:

• For $x = -1$:
  • $f(-1) = (-1)^2 – (-1) = 1 + 1 = 2$
  • $g(-1) = 2|-1 – 0.5| – 1 = 2|-1.5| – 1 = 2(1.5) – 1 = 2$
  • Match: Yes
• For $x = 0$:
  • $f(0) = 0^2 – 0 = 0$
  • $g(0) = 2|0 – 0.5| – 1 = 2(0.5) – 1 = 0$
  • Match: Yes
• For $x = 1$:
  • $f(1) = 1^2 – 1 = 0$
  • $g(1) = 2|1 – 0.5| – 1 = 2(0.5) – 1 = 0$
  • Match: Yes
• For $x = 2$:
  • $f(2) = 2^2 – 2 = 2$
  • $g(2) = 2|2 – 0.5| – 1 = 2(1.5) – 1 = 2$
  • Match: Yes

1. Reflexive: Must contain $\{(1,1), (2,2), (3,3)\}$.
2. Symmetric: Since it has $(1,2)$ and $(1,3)$, it must also have $\{(2,1), (3,1)\}$.
3. Current Set $R$: $\{(1,1), (2,2), (3,3), (1,2), (2,1), (1,3), (3,1)\}$.
4. Check Transitivity: $(2,1) \in R$ and $(1,3) \in R$. For transitivity, $(2,3)$ must be in $R$. It is not. So $R$ is not transitive.
5. Can we add more? If we add $(2,3)$, we must add $(3,2)$ for symmetry. The relation then becomes the Universal Relation ($A \times A$), which is transitive.

Thus, only the set constructed in step 3 satisfies the condition.

1. Smallest Equivalence Relation: Must be reflexive $\{(1,1), (2,2), (3,3)\}$ and symmetric for $(1,2)$, so add $\{(2,1)\}$.
   Set $R_1 = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$.
   This is transitive. (Equivalence class: $\{1, 2\}, \{3\}$).
2. Larger Relations: If we add any other element, say $(1,3)$, we must add $(3,1)$ (symmetry).
   Due to transitivity ($(2,1)$ and $(1,3)$ exist), we must add $(2,3)$.
   Due to symmetry for $(2,3)$, we must add $(3,2)$.
   This results in the Universal Relation $R_2 = A \times A$.

Only $R_1$ and $R_2$ are possible.